Summary Statistics for a Quantitative Variable
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AP Statistics › Summary Statistics for a Quantitative Variable
A wildlife biologist summarized the lengths (in centimeters) of 65 fish caught in a lake. The summaries were: mean $=24.5$ cm, median $=24.0$ cm, SD $=4.0$ cm, $Q_1=22.0$ cm, $Q_3=27.0$ cm, minimum $=15$ cm, maximum $=33$ cm (five-number summary: $15, 22, 24, 27, 33$). Which interpretation is correct?
Because the median is 24 cm, about half the fish are between 15 cm and 24 cm.
Since $Q_1=22$ cm, 25% of fish are longer than 22 cm.
The IQR is $33-15=18$ cm, so the middle 50% of fish lengths span 18 cm.
A typical fish length is $4.0$ cm because the SD equals the typical length.
The middle 50% of fish lengths are between 22 cm and 27 cm, so the IQR is 5 cm, and the distribution appears roughly symmetric because the mean and median are close and the tails are similar.
Explanation
This AP Statistics question tests interpreting quantitative variable summaries, including spread and shape. Choice C correctly notes middle 50% from 22 to 27 (IQR=5) and rough symmetry as mean (24.5) ≈ median (24.0) with similar tails (15-22=7, 27-33=6). Distractor E misstates Q1 (22), saying 25% longer than 22, but actually 75% ≥22. Lesson: Close mean-median suggests symmetry; IQR for central spread. SD (4.0) indicates low variability. Five-number summary shows balanced tails. Distribution appears nearly symmetric with moderate spread.
A hospital summarized the waiting time (in minutes) for 75 patients in an urgent care clinic. The summaries were: mean $=58.0$, median $=49.0$, SD $=30.0$, $Q_1=32.0$, $Q_3=70.0$, minimum $=8$, maximum $=160$ (five-number summary: $8, 32, 49, 70, 160$). Which interpretation is correct?
Because $Q_1=32$, 25% of patients waited less than 8 minutes.
The IQR equals $160-8=152$ minutes, so half of waits vary by 152 minutes.
A typical waiting time is 30 minutes because that is the SD.
The distribution is likely right-skewed, with a typical wait near 49 minutes, because the mean exceeds the median and the upper tail is much longer.
The middle 50% of waiting times are between 8 and 160 minutes because those are the minimum and maximum.
Explanation
In AP Statistics, this tests understanding quantitative variable summaries, emphasizing shape and center. Choice B correctly describes right-skewness with typical wait near 49 (median), as mean (58) > median and upper tail (70 to 160) much longer than lower (8 to 32). Distractor A confuses middle 50% with min-max (8 to 160) instead of Q1-Q3. Key: For skew, median is robust center; compare tails in five-number summary. IQR (70-32=38) measures central spread. SD (30) shows high variability. Indicates right-skewed waits with long upper extremes.
A wildlife biologist measured the mass (in grams) of 55 fish caught in a lake. The summary statistics were: mean $= 410$ g, median $= 395$ g, standard deviation $= 120$ g, five-number summary $(150, 320, 395, 470, 920)$, and IQR $= 150$ g. Which interpretation is correct?
Since $Q_3=470$ g, about 75% of fish weigh more than 470 g.
Because the median is 395 g, exactly half the fish weigh 395 g.
Because the maximum is much larger than $Q_3$ and the mean is greater than the median, the distribution is likely skewed right due to a few very large fish.
The standard deviation of 120 g means the typical fish weighs 120 g.
The IQR of 150 g means the heaviest fish is 150 g heavier than the lightest fish.
Explanation
This question assesses recognition of right-skewed distributions in biological data. The mean (410 g) being greater than the median (395 g) suggests right skew, where the distribution has a tail extending toward heavier fish. The maximum value (920 g) being much larger than Q3 (470 g) strongly confirms this - there are a few unusually large fish pulling the mean upward. The distractors contain typical errors: IQR represents Q3-Q1, not max-min; standard deviation measures variability around the mean, not a typical weight; Q3=470 g means 75% of fish weigh 470 g or less, not more; and the median indicates that half the fish weigh less than and half weigh more than 395 g, not that half weigh exactly 395 g.
A city collected the daily number of emergency-room visits for 45 days. The summary statistics were: mean $= 112$ visits, median $= 105$ visits, standard deviation $= 26$ visits, five-number summary $(60, 92, 105, 125, 210)$, and IQR $= 33$ visits. Which interpretation is correct?
The standard deviation of 26 visits means the range of the data is 26 visits.
An IQR of 33 visits means the number of visits typically equals 33.
Because $Q_1=92$ and $Q_3=125$, about 75% of days had between 92 and 125 visits.
Because the mean is greater than the median and the maximum is far above $Q_3$, the distribution is likely skewed right due to a few unusually high-visit days.
Since the median is 105, exactly 105 visits occurred on half of the days.
Explanation
This question assesses recognition of right-skewed distributions through summary statistics. The mean (112 visits) exceeding the median (105 visits) is a classic indicator of right skew, where the distribution has a tail extending toward higher values. Additionally, the maximum value (210) being far above Q3 (125) confirms this interpretation - there are a few days with unusually high emergency room visits pulling the mean upward. The incorrect options reflect common misunderstandings: IQR represents the spread of the middle 50% of data, not a typical value; standard deviation measures variability, not range; the median tells us that half the days had fewer than 105 visits, not that exactly 105 visits occurred; and the middle 50% of days (not 75%) had between Q1 and Q3 visits.
A delivery service recorded the distance (in miles) of 65 delivery routes in one day. The summary statistics were: mean $= 18.1$ mi, median $= 18.0$ mi, standard deviation $= 4.2$ mi, five-number summary $(8.5, 15.0, 18.0, 21.0, 27.5)$, and IQR $= 6.0$ mi. Which interpretation is correct?
The standard deviation of 4.2 miles means the typical route distance is 4.2 miles.
The IQR of 6.0 miles means the range of route distances is 6.0 miles.
Because the median is 18.0 miles, exactly half the routes are 18.0 miles long.
Because the mean and median are essentially equal and the five-number summary is fairly balanced, the distribution is likely roughly symmetric.
Since $Q_3=21.0$ miles, about 75% of routes are longer than 21.0 miles.
Explanation
This question tests recognition of symmetric distributions through summary statistics. The mean (18.1 miles) and median (18.0 miles) being essentially equal is a strong indicator of a roughly symmetric distribution. The five-number summary also shows good balance - the distances from median to quartiles are equal (3.0 miles each), and the extremes are roughly equidistant from their nearest quartiles. The incorrect options demonstrate common errors: IQR represents the spread of the middle 50% of data, not the total range; standard deviation measures typical deviation from the mean, not a typical value itself; Q3=21.0 means 75% of routes are 21.0 miles or shorter, not longer; and the median indicates that half the routes are shorter and half are longer than 18.0 miles, not that half are exactly 18.0 miles.
A library recorded the number of books checked out per visit for 75 patrons on a weekday. The summaries were: mean = 3.6 books, median = 3 books, SD = 2.2 books, five-number summary = (min 0, $Q_1$ 2, median 3, $Q_3$ 5, max 12), and IQR = 3. Which interpretation is correct?
Because the IQR is 3, the range is 3 books (from 0 to 3).
Because the median is 3, about 75% of patrons checked out fewer than 3 books.
Because the SD is 2.2, most patrons checked out about 2.2 books.
Because the IQR is 3, the number of books checked out ranged from 2 to 5 books.
Because the mean is slightly larger than the median and the maximum is well above $Q_3$, the distribution is likely right-skewed, with a few patrons checking out many books.
Explanation
This question examines mild right skewness in count data. The mean (3.6 books) being slightly larger than the median (3 books) suggests right skewness, confirmed by the maximum (12 books) being well above Q3 (5 books). This indicates a few patrons checking out many books pull the mean upward. The IQR of 3 correctly measures the spread between Q1=2 and Q3=5, representing the middle 50% of the data. The standard deviation of 2.2 books measures variability, not that patrons typically checked out 2.2 books. The range is actually 12 books (from 0 to 12), not 3 books. Understanding these distinctions helps correctly interpret summary statistics for count data.
A city measured the commute time (in minutes) for 60 randomly selected workers. The summaries were: mean = 31.2, median = 30, SD = 8.5, five-number summary = (min 14, $Q_1$ 25, median 30, $Q_3$ 36, max 49), and IQR = 11. Which interpretation is correct?
Because the maximum is 49 minutes, at least half the commutes are longer than 49 minutes.
Because the SD is 8.5 minutes, the typical commute time is 8.5 minutes.
Because the median is 30 minutes, about 75% of workers commute less than 30 minutes.
Because the mean is close to the median and the tails are similar in length, the distribution is likely roughly symmetric, with the middle 50% between 25 and 36 minutes.
Because the IQR is 11 minutes, the range of commute times is 11 minutes.
Explanation
This question tests recognition of a roughly symmetric distribution. The mean (31.2) and median (30) are very close, suggesting symmetry. The five-number summary shows similar tail lengths: the minimum (14) is 16 minutes below the median, while the maximum (49) is 19 minutes above it. The IQR of 11 minutes correctly identifies that the middle 50% of commute times fall between Q1=25 and Q3=36 minutes. The standard deviation of 8.5 minutes measures spread, not a typical commute time. Remember that the median divides the data in half - 50% below and 50% above, not 75% below as one distractor suggests.
An online retailer summarized the delivery times (in days) for 100 randomly selected orders. The summaries were: mean $=5.6$ days, median $=5.0$ days, SD $=2.4$ days, $Q_1=4.0$ days, $Q_3=6.0$ days, minimum $=2$ days, maximum $=18$ days (five-number summary: $2, 4, 5, 6, 18$). Which interpretation is correct?
Because the mean is greater than the median, at least 50% of orders took more than 5.6 days.
The IQR is $6.0-4.0=2.0$ days, meaning the middle 50% of delivery times span about 2 days.
The distribution is likely symmetric because $Q_1$ and $Q_3$ are equally spaced from the median.
A typical order took $2.4$ days to arrive because the SD is $2.4$ days.
The IQR is $18-2=16$ days, which shows the middle 50% is very spread out.
Explanation
This AP Statistics question evaluates summary statistics for quantitative variables, particularly spread. Choice B accurately computes IQR as 6-4=2 days, spanning middle 50%. Distractor D miscalculates IQR as range (18-2=16), ignoring actual quartiles. Lesson: IQR resists outliers, ideal for skew; here mean (5.6) > median (5.0) suggests right-skew despite close Q1/median/Q3. SD (2.4) quantifies deviation, not typical time. Five-number summary shows long upper tail. Distribution is right-skewed with low central spread but outliers.
A track coach recorded the times (in seconds) for 40 athletes to complete a 400-meter run. The summaries were: mean $=62.1$ s, median $=61.8$ s, standard deviation $=2.3$ s, five-number summary (min $=57.5$, $Q_1=60.6$, median $=61.8$, $Q_3=63.4$, max $=67.0$), and IQR $=2.8$ s. Which interpretation is correct?
The IQR is $67.0-57.5=9.5$ seconds because IQR is the same as the range.
Since the maximum is $67.0$ s, at least 75% of athletes ran slower than $67.0$ s.
Because the mean is slightly greater than the median, the distribution must be extremely right-skewed.
The standard deviation of $2.3$ seconds means most athletes ran exactly $2.3$ seconds.
The middle 50% of times are between $60.6$ s and $63.4$ s, so the IQR is $2.8$ s.
Explanation
This question assesses IQR and quartile interpretation for run times. Q1 = 60.6 s and Q3 = 63.4 s mean middle 50% between them, with IQR = 2.8 s, as choice A states. Choice D is a distractor, confusing IQR with range (9.5 s). Mini-lesson: IQR = Q3 - Q1 measures central spread; quartiles divide data into quarters. Slight mean (62.1) > median (61.8) suggests mild right skew, and standard deviation (2.3 s) quantifies deviation from mean, not implying 'most' values are exactly that.
A school nurse recorded the number of minutes 60 students slept the night before a standardized test. The summary statistics were: mean $= 420$ min, median $= 435$ min, standard deviation $= 55$ min, five-number summary $(
\min, Q_1, \text{Med}, Q_3, \max) = (290, 390, 435, 465, 520)$, and IQR $= 75$ min. Which interpretation is correct?
The IQR of 75 minutes means the range of sleep times is 75 minutes.
The standard deviation of 55 minutes means most students slept about 55 minutes total.
The median of 435 minutes means exactly half the students slept 435 minutes.
Since $Q_3=465$ and $Q_1=390$, about 75% of students slept between 390 and 465 minutes.
Because the mean is less than the median, the distribution is likely skewed left, possibly due to a few students with very low sleep times.
Explanation
This question tests understanding of how the relationship between mean and median indicates skewness in a distribution. When the mean (420 minutes) is less than the median (435 minutes), this suggests the distribution is skewed left, with a tail extending toward lower values. This occurs because the mean is pulled down by a few students who slept much less than typical. The other options contain common misconceptions: IQR measures the spread of the middle 50% of data, not the total range; standard deviation measures typical deviation from the mean, not total sleep time; the middle 50% of students slept between Q1 and Q3, not 75%; and the median indicates that half slept less than and half slept more than 435 minutes, not that half slept exactly 435 minutes.